ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$Sym ({1, 2, 3})$ has exactly six elements, is non-abelian, and its elements have orders $1$ , $2$ and $3$

Example

Let $X = {1, 2, 3}$ , the three-element set of natural numbers $1 = σ (0)$ , $2 = σ (1)$ , $3 = σ (2)$ , and let $G = Sym (X)$ be its symmetric group (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition), a group under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements). Then:

$G$ has exactly six elements, namely

$i d, (1 2), (1 3), (2 3), (1 2 3), (1 3 2),$

so $G \approx 6$ and $∣ G ∣ = 6$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity);
$G$ is not abelian: $(1 2) \circ (2 3) = (1 2 3)$ while $(2 3) \circ (1 2) = (1 3 2)$ ;
the orders of its elements are $ord (i d) = 1$ , $ord = 2$ for each of the three transpositions, and $ord = 3$ for each of the two $3$ -cycles.

Facts & Assumptions

Given: $X = {1, 2, 3}$ and $G = Sym (X)$ with the operation $f \circ g$ , $(f \circ g) (x) = f (g (x))$ , and identity $i d$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L1]

$1$ , $2$ and $3$ are pairwise distinct natural numbers: $2 = σ (1) = 1 \cup {1}$ and $3 = σ (2) = 2 \cup {2} = {0, 1, 2}$ , so $1 \in 2$ , $2 \in 3$ and $1 \in 3$ , while no natural number is a member of itself (The natural numbers $N$ (von Neumann), Every natural number is a transitive set and is not a member of itself); so $X$ has exactly three elements.

[L2]

$Sym (X)$ is a group under composition; the cycle symbols $(a b)$ and $(a b c)$ denote the permutations described in The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition ( $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Group and abelian group).

[L3]

Two functions $X \to X$ are equal exactly when they agree at every point; a function on $X$ is determined by the triple $(f (1), f (2), f (3))$ , and $f$ is a bijection of $X$ exactly when those three values are pairwise distinct, since three distinct values in a three-element set exhaust it (Injection, surjection, bijection).

[L4]

Powers: $f^{0} = i d$ , $f^{1} = f$ and $f^{σ (k)} = f^{k} \circ f$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ); $f^{j + k} = f^{j} \circ f^{k}$ and $(f^{j})^{k} = f^{j k}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute); $ord (f)$ is the least $k \geq 1$ with $f^{k} = i d$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L5]

Finiteness and counting: $A \approx m$ means a bijection $A \to m$ exists, and $∣ A ∣ = m$ is then the unique such natural (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L6]

If $ord (f) = k$ then $f^{j} \neq i d$ for $1 \leq j < k$ , and conversely a $k$ with $f^{k} = i d$ and $f^{j} \neq i d$ for $1 \leq j < k$ is the order (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Verification

technique · direct

1.1

Writing each element of $G$ as the triple of its values at $1, 2, 3$ , the six listed permutations are $i d = (1, 2, 3)$ , $(2 3) = (1, 3, 2)$ , $(1 2) = (2, 1, 3)$ , $(1 2 3) = (2, 3, 1)$ , $(1 3 2) = (3, 1, 2)$ and $(1 3) = (3, 2, 1)$ . These six triples are pairwise different, so the six permutations are pairwise different.

L1L2L3

1.2

Each transposition $τ \in {(1 2), (1 3), (2 3)}$ satisfies $τ \circ τ = i d$ , since it exchanges two points and fixes the third, so applying it twice returns every point to itself; and $τ \neq i d$ , since it moves two points.

L2L3

1.3

$ord (i d) = 1$ , since ${i d}^{1} = i d$ and $1$ is the least natural that is $\geq 1$ .

L4L6

2.1

Every element of $G$ is one of the six. Let $f \in G$ ; its triple has pairwise distinct entries by [L3]. There are three possible values for $f (1)$ ; for each, two remaining values for $f (2)$ ; and then $f (3)$ is forced to be the one element of $X$ left over. Running through those six combinations produces exactly the six triples listed in step 1.1.

step 1.1L1L3

2.2

Composites, computed pointwise. $(1 2) \circ (2 3)$ sends $1 \mapsto 1 \mapsto 2$ , $2 \mapsto 3 \mapsto 3$ , $3 \mapsto 2 \mapsto 1$ , giving the triple $(2, 3, 1)$ , which is $(1 2 3)$ . And $(2 3) \circ (1 2)$ sends $1 \mapsto 2 \mapsto 3$ , $2 \mapsto 1 \mapsto 1$ , $3 \mapsto 3 \mapsto 2$ , giving $(3, 1, 2)$ , which is $(1 3 2)$ .

step 1.1L2L3

2.3

Let $c = (1 2 3)$ , so $c$ sends $1 \mapsto 2 \mapsto 3 \mapsto 1$ . Then $c \circ c$ sends $1 \mapsto 3$ , $2 \mapsto 1$ , $3 \mapsto 2$ , which is the triple $(3, 1, 2)$ , that is $(1 3 2)$ ; and $c \circ c \circ c$ sends $1 \mapsto c (3) = 1$ , $2 \mapsto c (1) = 2$ , $3 \mapsto c (2) = 3$ , so it is $i d$ .

step 1.1L2L3

2.4

Each transposition $τ$ has $ord (τ) = 2$ : $τ^{2} = τ \circ τ = i d$ by step 1.2, while $τ^{1} = τ \neq i d$ , so $2$ is the least $k \geq 1$ with $τ^{k} = i d$ .

step 1.2L4L6

3.1

By steps 1.1 and 2.1 the set $G$ consists of exactly the six listed elements, so the map $6 \to G$ sending $0, 1, 2, 3, 4, 5$ to $i d, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)$ respectively is a bijection; hence $G \approx 6$ and $∣ G ∣ = 6$ . This is claim 1.

step 1.1step 2.1L5

3.2

By step 2.2 the two composites differ, since the triples $(2, 3, 1)$ and $(3, 1, 2)$ differ at $1$ ; so $G$ is not abelian, which is claim 2.

step 2.2L3

3.3

Each of $c = (1 2 3)$ and $c^{2} = (1 3 2)$ has order $3$ : by step 2.3, $c^{3} = i d$ , $c^{1} = c \neq i d$ and $c^{2} = (1 3 2) \neq i d$ , so $ord (c) = 3$ ; and $(c^{2})^{3} = c^{6} = i d$ with $c^{2} \neq i d$ and $(c^{2})^{2} = c^{4} = c \neq i d$ , so $ord (c^{2}) = 3$ as well.

step 2.3L4L6

4.1

Claims 1, 2 and 3 are steps 3.1, 3.2 and steps 1.3, 2.4 and 3.3 taken together; the orders occurring are exactly $1$ , $2$ and $3$ .

step 3.1step 3.2step 1.3step 2.4step 3.3∎

Remarks

This is the first explicitly non-abelian group in the library. Claim 2 instantiates the general statement of $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements at the three distinct points $1, 2, 3$ ; the pair of transpositions used there is the pair used here.
No element of $G$ has order $6$ , by claim 3, so by If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ every cyclic subgroup $⟨ f ⟩$ has at most three elements and none of them is all of $G$ : the group is not cyclic. That also follows from claim 2, every cyclic group being abelian ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).
The elements $1$ , $2$ , $3$ here are the natural numbers of The natural numbers $N$ (von Neumann), hence sets; nothing in the computation uses anything about them beyond their being three distinct objects.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Symmetric group (Wikipedia) (standard reference, not scraped)
Dihedral group of order 6 (Wikipedia) (standard reference, not scraped)

Sym⁡({1,2,3})Sym({1,2,3}) has exactly six elements, is non-abelian, and its elements have orders 11, 22 and 33